# 19.1: Oxidation-Reduction Reactions

Hypothes.is Tag = c1403c19 (case sensitive)
Group = UALRChem1403

## Introduction

Oxidation-Reduction (Redox) reactions are a basic type of chemical reaction involving the transfer of electrons from one atom or chemical entity to another, or more accurately, from one type of orbital to another, that results in new bonds being formed. These were introduced in gen chem 1, section 3.6 Redox Reactions, and you may want to quickly review that material now, and be prepared to return to it if you need help in identifying the oxidation state of an atom. Note, the electron transfer may be between two different chemical species, or between atoms of the same species. This chapter introduces electrochemistry, which is the study of redox reactions.

### Redox Review

Consider the reaction of elemental zinc and sulfur to form zinc sulfide.

Figure $$\PageIndex{1}$$: Reaction between zinc and sulfur.

Zinc is losing electrons in the reaction and is thus oxidized to the zinc cation, while sulfur is gaining electrons and is thus reduced to the sulfide anion.

\begin{align} \textbf{oxidation}&=\textrm{loss of electrons}\\ \textbf{reduction}&=\textrm{gain of electrons} \end{align}

Note, we often use the term oxidant and reductant to introduce the species that cause oxidation and reduction, and thus zinc is the reducing agent (causes reduction, while being oxidized), and sulfur is the oxidizing agent (causes oxidation, while being reduced).

\begin{align} \textbf{Reductant (reducing agent)}&=\textrm{species that is oxidized}\\ \textbf{Oxidant (oxidizing agent)}&=\textrm{species that is reduced} \end{align}

In the above example the electron(s) were being transferred from one chemical entity to another. When this happens we can split the chemical reaction into two half-reactions, one for the oxidation and one for the reduction. Note, these are coupled by the electrons transferred, and it is important to realize that the two half-reactions occur concurrently (the electrons leave the zinc and go to the sulfur). It should be noted that not all redox reactions can be broken into simple half-reactions, as sometimes the atoms being oxidized and those being reduced are different atoms on the same compound.

$\underbrace{Zn→Zn^{2+}+2e^−}_{\text{Oxidation}}$

$\underbrace{S+ 2 e^−→S^{2−}}_ {\text{Reduction}}$

However, not all redox reactions involve the transfer of electrons to produce ions. Consider, for example the standard formation reaction to form hydrogen sulfide from sulfur and hydrogen, which is similar to the above reaction for the production of zinc sulfide, but hydrogen sulfide is a covalent bond.

$\ce{H2}+\ce{S}\rightarrow \ce{H_2S}$

Because hydrogen sulfide is a covalent compound, electron transfer from the reductant to the oxidant does not occur in the explicit sense. But a look at the polarity of the covalent bond shows that the more electronegative sulfur has a higher density, and so electron density has been transferred. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. Review the guidelines (3.6.2.1) from gen chem 1 on assigning oxidation numbers to each element in a molecule or ion.

Using the oxidation number provides a better and more universal definition of oxidation and reduction.

\begin{align} \textbf{oxidation}&=\textrm{increase in oxidation number}\\ \textbf{reduction}&=\textrm{decrease in oxidation number} \end{align}

## Balancing Redox Reactions

A balanced chemical equation must not only conserve mass and energy, but also charge. Consider the following equation for the formation of Iodine and ferrous ion from iodide and ferric ions, is it balanced?
$Fe^{+3}+2I^{-} \rightarrow Fe^{+2} + I_2$.

Sure mass is balanced, but charge is not, as it is +1 on the reactant side and +2 on the product, which means an electron disappeared as products were being formed. Taking a closer look from the half-reaction perspective that two electrons were released but only one was transferred, and that an electron just disappeared.

$\underbrace{2I^-→I_2+2e^−}_{\text{Oxidation}}$

$\underbrace{Fe^{+3}+ 1e^−→Fe^{+2}}_ {\text{Reduction}}$

For the reaction to be balanced the number of electrons released in the oxidation half-reaction must equal the number gained in the reduction half-reaction, which means everything in the reduction half-reaction needs to be multiplied by two.

$\underbrace{2I^-→I_2+2e^−}_{\text{Oxidation}}$

$\underbrace{2Fe^{+3}+ 2e^−→2Fe^{+2}}_ {\text{Reduction}}$

Note the two half-reactions are coupled by the electrons transferred, and the balanced equation would be the sum of the above two equations

$2Fe^{+3}+2I^{-} \rightarrow 2Fe^{+2}+ I_2$.

Note, charge is conserved in the above equation, which has a charge of +4 on both sides.

There are many techniques for balancing redox reactions and we will introduce two. The first is a half-reaction method that does not require you to calculate the oxidation number and can be used when the species being oxidized is a completely different chemical entity than the one being reduced. The second technique requires calculating the oxidation number and is needed for problems that can not be broken up into half-reactions, like ones where different atoms on the same species are getting oxidized and reduced, that is, electrons are being transferred across atoms within a molecule or compound.

### Half-Reaction Method

If the chemical species being oxidized and reduced are different entities they can be separated into half-reactions, balanced for mass and charge, and then added in a way that balances the total charge. Many electrochemical systems are in aqueous solutions, with many also being in acidic or basic environments (an alkaline battery is a basic environment, and your car battery has battery acid, making it an acidic environment), So in this section we will learn how to balance reactions in acidic and basic environments.

#### Balancing Acidic Redox Reactions

The following steps allow you to balance a redox reaction and you often do not even have to identify the oxidation state of the various species.

1. Split Skeletal Eq. into 1/2 Reactions
2. For Each Half-Reaction
1. Balance all Elements Except O and H
2. Balance O by adding water
3. Balance H by adding H+.
4. Balance Charge by adding e-.
3. Multiply 1/2 rxns by appropriate integer so electrons lost = electrons gained.
4. Add half-reactions, cancelling electrons lost and gained, and check work

#### Balancing Basic Redox Reactions

Balance as if it was acidic, and then add OH- to both sides to neutralize H+ (converting it to H2O)

1. Split Skeletal Eq. into 1/2 Reactions
2. For Each Hal- Reaction
1. Balance all Elements Except O and H
2. Balance O by adding water
3. Balance H by adding H+.
4. Balance Charge by adding e-.
3. Multiply 1/2 rxns by appropriate integer so electrons lost = electrons gained.
4. Add half-reactions, cancelling electrons lost and gained, and check work
5. Add OH- to both sides to neutralize H+ (converting it to H2O)

(only the last step is different from balancing acidic reactions)

Exercise $$\PageIndex{1}$$

Balance the reaction of thiosulfate with iodine to produce sulfate and iodide in basic conditions.

$\ce{S_2O_3^{-2} + I_2 \rightarrow SO_4^{-2} + I^{-}} \nonumber$

The following video goes over the solution to this problem, which is:

Answer: $10 OH^- +S_2O_3^{-2} + 4I_2 \leftarrow 2SO_4^{-2} + 5H_2O + 8I^-$

### Oxidation Method

This method can be used in reactions where one chemical has atoms being oxidized and other atoms being reduced, that is, the chemical is in both half-reactions.

1. Identify oxidation number of all atoms being oxidized or reduced.
• Indicate oxidation state above the atom
• Review section 3.6.2.1 on oxidation numbers if needed
2. Balance elements in "coupled" molecules
• Balance elements other than oxygen or hydrogen.
3. Use a line to connect atoms undergoing a change in oxidation number:
• Above Eq. for atoms being oxidized
• Below Eq. for atoms being reduced
4. Balance Oxidizing and Reducing Equivalents
• multiply by appropriate coefficients so total number on top equals total on bottom
5. If aqueous balance Oxygen by adding water
1. If acidic balance hydrogen by adding H+.
2. If basic, treat as acidic and then add OH- to remove H+.
6. Double check work.

## Contributors

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: